Analytic Geometry — Circle (Basic)

Circle equation: \((x-a)^2 + (y-b)^2 = r^2\)
The centre is \((a, b) = (3, 2)\)

From the equation: \(r^2 = 16\) therefore \(r = \sqrt{16} = 4\)

Substitute the point into the equation: \((6-3)^2 + (2-2)^2 = 9 + 0 = 9 = r^2\)
The result equals 9, so yes — the point is on the circle!

Circle with centre at the origin: \(x^2 + y^2 = r^2 = 5^2 = 25\)

Equation: \((x-2)^2 + (y-(-3))^2 = 4^2\)
i.e.: \((x-2)^2 + (y+3)^2 = 16\)

Distance between points: \(d = \sqrt{(5-2)^2 + (1-5)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

\(r^2 = 36\) therefore \(r = \sqrt{36} = 6\)

Substituting: \(3^2 + 4^2 = 9 + 16 = 25 ✓\)
Yes! 3-4-5 triangle — the point is on the circle

\((x+1)^2 = (x-(-1))^2\)
The centre: \((-1, 4)\)

Radius = distance from centre to point:
\(r = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9+16} = \sqrt{25} = 5\)

\(r^2 = 49\) therefore \(r = \sqrt{49} = 7\)

Substituting: \(0^2 + 5^2 = 0 + 25 = 25 ✓\)
Yes! The point is on the y-axis at distance 5 from the origin

Equation: \((x-(-2))^2 + (y-3)^2 = 6^2\)
i.e.: \((x+2)^2 + (y-3)^2 = 36\)

An equation of the form \(x^2 + y^2 = r^2\) — the centre is always at the origin \((0,0)\)

Distance from (0,0) to centre (3,4): \(d = \sqrt{3^2 + 4^2} = 5\)
The distance equals the radius, yes!

\(r^2 = 64\) therefore \(r = \sqrt{64} = 8\)

Radius = distance:
\(r = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5\)

From the equation: \((x-7)^2 + (y-(-4))^2\)
The centre: \((7, -4)\)

Substituting: \(6^2 + 8^2 = 36 + 64 = 100 ✓\)
Yes! 6-8-10 triangle

Equation: \((x-5)^2 + (y-(-1))^2 = 3^2\)
i.e.: \((x-5)^2 + (y+1)^2 = 9\)

On the x-axis: \(y=0\)
Substituting: \(x^2 + 0 = 16\) → \(x = ±4\)
Points: \((4,0), (-4,0)\)

On the y-axis: \(x=0\)
Substituting: \(0 + y^2 = 25\) → \(y = ±5\)
Points: \((0,5), (0,-5)\)

On the y-axis: \(x=0\)
Substituting: \((0-3)^2 + y^2 = 4\)
\(9 + y^2 = 4\) → \(y^2 = -5\)
No solution! The circle is too far from the y-axis

On the x-axis: \(y=0\)
Substituting: \((x-2)^2+(0-3)^2=9\)
\((x-2)^2+9=9\) → \((x-2)^2=0\)
No real crossing — tangency only

On the x-axis: \(y=0\)
Substituting: \(x^2 + 16 = 16\) → \(x^2 = 0\) → \(x = 0\)
Tangency at one point — but there is an "intersection"

With x-axis: \((3,0), (-3,0)\) - 2 points
With y-axis: \((0,3), (0,-3)\) - 2 points
Total: 4 points

Centre on x-axis, radius 5
Intersects the x-axis at \((0,0)\) and \((10,0)\)
Does not intersect the y-axis (too far)

Substitute \(y=3\) into the circle equation:
\(x^2 + 9 = 25\) → \(x^2 = 16\) → \(x = ±4\)
2 Points: \((4,3), (-4,3)\)

Substituting \(x=5\):
\(25 + y^2 = 16\) → \(y^2 = -9\)
No solution! The line is too far away (radius=4, line at x=5)

Substituting \(y=x\):
\(x^2 + x^2 = 8\) → \(2x^2 = 8\) → \(x^2 = 4\) → \(x = ±2\)
Points: \((2,2), (-2,-2)\)

The line \(y=4\) passes through the centre of the circle \((3,4)\)
Substituting: \((x-3)^2 + 0 = 25\) → \(x-3 = ±5\)
Points: \((8,4), (-2,4)\)

Substituting \(x=0\):
\(1 + y^2 = 1\) → \(y^2 = 0\) → \(y = 0\)
One point: \((0,0)\) - Tangent!

In a right triangle: the circumcenter is the midpoint of the hypotenuse
Radius = half the hypotenuse = \(\frac{10}{2} = 5\)

The hypotenuse is 5 (the longest side)
radius = \(\frac{5}{2} = 2.5\)

Hypotenuse = diameter of the circle = \(2r = 2 \cdot 6 = 12\)

The right angle is at A. The hypotenuse is BC
Circumcenter = midpoint of hypotenuse = \((\frac{8+0}{2}, \frac{0+6}{2}) = (4,3)\)

Hypotenuse = 13, radius = \(\frac{13}{2} = 6.5\)
Equation: \(x^2 + y^2 = 6.5^2 = 42.25\)

From the equation: \(r^2 = 100\) → \(r = 10\)
Hypotenuse = diameter = \(2r = 20\)

In a right triangle, the circumcenter is always at the midpoint of the hypotenuse (the longest side)

Rectangle diagonal: \(d = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10\)
Circle radius = half the diagonal = \(\frac{10}{2} = 5\)

Square diagonal: \(d = 10\sqrt{2}\)
Radius = half the diagonal = \(\frac{10\sqrt{2}}{2} = 5\sqrt{2}\)

Radius = 13 → diameter = 26
Rectangle diagonal: \(\sqrt{10^2 + b^2} = 26\)
\(100 + b^2 = 676\) → \(b^2 = 576\) → \(b = 24\)

\(r^2 = 50\) → \(r = 5\sqrt{2}\)
diagonal = \(2r = 10\sqrt{2}\)
side = \(\frac{10\sqrt{2}}{\sqrt{2}} = 10\)

Rectangle sides: 6 and 8
Diagonal: \(\sqrt{6^2+8^2} = 10\)
Radius: \(r = 5\)
Equation: \(x^2 + y^2 = 25\)

Circumradius formula: \(R = \frac{abc}{4S}\)
\(R = \frac{5 \cdot 6 \cdot 7}{4 \cdot 14.7} = \frac{210}{58.8} \approx 3.57\)

\(R = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{2730}{336} = 8.125\)

In an equilateral triangle: \(R = \frac{a}{\sqrt{3}}\)
\(R = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\)

Sides: AB=6, AC=5, BC=5 (isosceles triangle)
Area = 12
\(R = \frac{6 \cdot 5 \cdot 5}{4 \cdot 12} = \frac{150}{48} = 3.125\)

The formula \(R = \frac{abc}{4S}\) requires individual side lengths, not just the perimeter.
Cannot be computed from perimeter and area alone.

Height: \(h = \sqrt{5^2 - 4^2} = 3\)
Area: \(S = \frac{8 \cdot 3}{2} = 12\)
\(R = \frac{5 \cdot 5 \cdot 8}{4 \cdot 12} = \frac{200}{48} = \frac{25}{6}\)

Height: \(h = \sqrt{10^2 - 6^2} = 8\)
Area: \(S = \frac{12 \cdot 8}{2} = 48\)
\(R = \frac{10 \cdot 10 \cdot 12}{4 \cdot 48} = \frac{1200}{192} = 6.25\)

Using the formula \(R = \frac{a \cdot a \cdot b}{4S}\) and area \(S = \frac{b \cdot h}{2}\)
Height: \(h = \sqrt{a^2 - 9}\)
After solving: \(a = 5\)

From area: \(30 = \frac{10 \cdot h}{2}\) → \(h = 6\)
Leg: \(a = \sqrt{5^2 + 6^2} = \sqrt{61}\)
\(R = \frac{\sqrt{61} \cdot \sqrt{61} \cdot 10}{4 \cdot 30} = \frac{610}{120} \approx 6.01\)

Height: \(h = 12\) (5-12-13 triangle)
Area: \(S = 60\)
\(R = \frac{13 \cdot 13 \cdot 10}{4 \cdot 60} = \frac{1690}{240} = 6.5\)
Equation: \(x^2 + y^2 = 6.5^2 = 42.25\)